Tree Cutting
题目连接：
http://acm.hdu.edu.cn/showproblem.php?pid=5909
Description
Byteasar has a tree T with n vertices conveniently labeled with 1,2,...,n. Each vertex of the tree has an integer value vi.
The value of a non-empty tree T is equal to v1⊕v2⊕...⊕vn, where ⊕ denotes bitwise-xor.
Now for every integer k from [0,m), please calculate the number of non-empty subtree of T which value are equal to k.
A subtree of T is a subgraph of T that is also a tree.
Input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, the first line of the input contains two integers n(n≤1000) and m(1≤m≤210), denoting the size of the tree T and the upper-bound of v.
The second line of the input contains n integers v1,v2,v3,...,vn(0≤vi<m), denoting the value of each node.
Each of the following n−1 lines contains two integers ai,bi, denoting an edge between vertices ai and bi(1≤ai,bi≤n).
It is guaranteed that m can be represent as 2k, where k is a non-negative integer.
Output
For each test case, print a line with m integers, the i-th number denotes the number of non-empty subtree of T which value are equal to i.
The answer is huge, so please module 109+7.
Sample Input
2
4 4
2 0 1 3
1 2
1 3
1 4
4 4
0 1 3 1
1 2
1 3
1 4
Sample Output
3 3 2 3
2 4 2 3
Hint
题意
给你一棵树，然后问你，里面有多少个连通块的异或和为j
题解：
树形DP去做，然后我们考虑转移，用fwt去进行优化就好了
至于为什么这么变换的。。。。
俺也不是很清楚：http://picks.logdown.com/posts/179290-fast-walsh-hadamard-transform
代码
#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int inv = (mod+1)/2;
const int maxn = 3005;
int n,m,ans[maxn],val[maxn];
vector<int> E[maxn];
long long dp[maxn][maxn],tmp[maxn],a[maxn],b[maxn];
void fwt(long long *aa,int l,int r)
{
if(r-l==1)return;
int mid=(l+r)/2;
fwt(aa,l,mid);
fwt(aa,mid,r);
int len=mid-l;
for(int i=l;i<mid;i++)
{
long long x1=aa[i];
long long x2=aa[i+len];
aa[i]=(x1+x2)%mod;
aa[i+len]=(x1-x2+mod)%mod;
}
}
void ifwt(long long *aa,int l,int r)
{
if(r-l==1)return;
int mid=(l+r)/2;
int len=mid-l;
for(int i=l;i<mid;i++)
{
long long y1=aa[i];
long long y2=aa[i+len];
aa[i]=(y1+y2)*inv%mod;
aa[i+len]=((y1-y2+mod)%mod*inv)%mod;
}
ifwt(aa,l,mid);
ifwt(aa,mid,r);
}
void solve(long long *aa,long long *bb)
{
memcpy(a,aa,sizeof(long long)*m);
memcpy(b,bb,sizeof(long long)*m);
fwt(a,0,m);
fwt(b,0,m);
memset(tmp,0,sizeof(tmp));
for(int i=0;i<m;i++)
tmp[i]=a[i]*b[i]%mod;
ifwt(tmp,0,m);
}
void dfs(int x,int f)
{
memset(dp[x],0,sizeof(dp[x]));
dp[x][val[x]]=1;
for(int i=0;i<E[x].size();i++){
int v=E[x][i];
if(v==f)continue;
dfs(v,x);
solve(dp[x],dp[v]);
for(int j=0;j<m;j++)(dp[x][j]+=tmp[j])%=mod;
}
for(int i=0;i<m;i++)(ans[i]+=dp[x][i])%=mod;
}
void solve()
{
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)ans[i]=0;
for(int i=1;i<=n;i++)scanf("%d",&val[i]),E[i].clear();
for(int i=1;i<n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
E[x].push_back(y);
E[y].push_back(x);
}
dfs(1,0);
for(int i=0;i<m;i++)
{
if(i==0)printf("%d",ans[i]);
else printf(" %d",ans[i]);
}
printf("
");
}
int main()
{
int t;scanf("%d",&t);
while(t--)solve();
return 0;
}